Theorem difdif 4130

Description: Double class difference. Exercise 11 of [TakeutiZaring] p. 22. (Contributed by NM, 17-May-1998.)

Assertion

Ref	Expression
difdif	⊢ (𝐴 ∖ (𝐵 ∖ 𝐴)) = 𝐴

Proof of Theorem difdif

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		pm4.45im 826	. . 3 ⊢ (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴)))
2		iman 400	. . . . 5 ⊢ ((𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴) ↔ ¬ (𝑥 ∈ 𝐵 ∧ ¬ 𝑥 ∈ 𝐴))
3		eldif 3957	. . . . 5 ⊢ (𝑥 ∈ (𝐵 ∖ 𝐴) ↔ (𝑥 ∈ 𝐵 ∧ ¬ 𝑥 ∈ 𝐴))
4	2, 3	xchbinxr 334	. . . 4 ⊢ ((𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴) ↔ ¬ 𝑥 ∈ (𝐵 ∖ 𝐴))
5	4	anbi2i 621	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴)) ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ (𝐵 ∖ 𝐴)))
6	1, 5	bitr2i 275	. 2 ⊢ ((𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ (𝐵 ∖ 𝐴)) ↔ 𝑥 ∈ 𝐴)
7	6	difeqri 4123	1 ⊢ (𝐴 ∖ (𝐵 ∖ 𝐴)) = 𝐴

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 394   = wceq 1534   ∈ wcel 2099   ∖ cdif 3944

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-ext 2697

This theorem depends on definitions:  df-bi 206  df-an 395  df-tru 1537  df-ex 1775  df-sb 2061  df-clab 2704  df-cleq 2718  df-clel 2803  df-v 3464  df-dif 3950

This theorem is referenced by:  dif0  4377  undifabs  4482